Let $f$ and $g$ be integrable functions on the measure space $(X, \mathcal M, \mu)$ with the property that
$$ \mu(\{f > t\} \triangle \{g > t\}) = 0 $$
for $\lambda$-a.e. $t \in \mathbb R$. Prove that $f = g\;$ $\mu$-almost everywhere.
Here, $\lambda$ is Lebesgue measure and $\triangle$ denotes symmetric difference.
My thoughts: I'm quite certain I should apply Fubini here to some $F: X \times \mathbb R \to \mathbb R$. But I'm not sure how to choose $F$ and, even if I did have an $F$ in hand, I don't know how to apply Fubini without some completeness or $\sigma$-finite condition on $(X, \mathcal M, \mu)$.
Any hints would be much appreciated.
No need for Fubini.
Let $D_t = f^{-1}(t,\infty) \triangle g^{-1}(t,\infty)$. Note that if $s<t$, then $D_t \subset D_s$. It follows that $\mu D_t = 0$ for all $t$ (not just ae. [$\lambda$]).
Let $q_n$ be an enumeration of $\mathbb{Q}$. Define $\Delta_n = f^{-1}(q_n,\infty) \cap g^{-1}(-\infty, q_n]$. Let $\Delta = \cup_n \Delta_n$. Then we have $\Delta = \{ x | f(x) > g(x) \}$. Also, note that $\Delta_n \subset D_{q_n}$.
Suppose $\mu \Delta>0$, then $\mu \Delta_n >0$ for some $n$, which contradicts $\mu D_{q_n} = 0$. Hence $\mu \Delta = 0$.
Exactly the same analysis applies with $f,g$ swapped, hence $f(x) = g(x)$ ae. [$\mu$].